Question #6216a

1 Answer
Ernest Z.
Jan 7, 2017

K_c = K_P = 0.019Kc=KP=0.019

Explanation:

For the equilibrium "H"_2 + "I"_2 ⇌ "2HI"H2+I2⇌2HI,

K_c = (["H"_2]["I"_2])/(["HI"]^2)Kc=[H2][I2][HI]2

If the values given are the equilibrium concentrations,

K_c = ((2.9 × 10^"-3")(1.7 × 10^"-3"))/(1.6 × 10^"-2")^2 = 0.019Kc=(2.9×10-3)(1.7×10-3)(1.6×10-2)2=0.019

K_P = K_c(RT)^(Δn),

where Δn is the change in the number of moles of gas.

Δn = "2 - 2 = 0"

∴ K_p = K_c(RT)^0 = K_c = 0.019

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